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sammy [17]
1 year ago
11

Brian irons \dfrac29 9 2 ​ start fraction, 2, divided by, 9, end fraction of his shirt in 3\dfrac353 5 3 ​ 3, start fraction, 3,

divided by, 5, end fraction minutes. Brian irons at a constant rate. At this rate, how much of his shirt does he iron each minute?
Mathematics
2 answers:
Artyom0805 [142]1 year ago
5 0
For this case we can make the following rule of three:
 2/9 ------> 3/5
 x ---------> 1
 Clearing x we have:
 x = (1 / (3/5)) * (2/9)
 Rewriting we have:
 x = (5/3) * (2/9)
 x = 10/27
 Answer:
 
he irons 10/27 of his shirt every minute
Darya [45]1 year ago
3 0

Answer:

5/81

Step-by-step explanation:

i hope this helped

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Estimate the value of 50.75 x 0.18<br> Estimate the value of 196 divided by 0.499
DIA [1.3K]
<span>This question is a simple one. To answer this question, you need to understand the description in the question and determine to multiply or divide the number.

The first problem would be:
50.75 x 0.18= 9.135
</span>If you need to estimate, 50.75 is near 50; 0.18 is near 0.2 or 1/5 so it would be: 50/0.2= 10<span>

The second problem would be:
196 / 0.499: 392.785 
If you need to estimate, 0.499 is near 0.5 then 196/0.5 would be 392</span>
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2 years ago
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Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,500 skat
BartSMP [9]
Let x be a random variable representing the number of skateboards produced
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Problem 5 (4+4+4=12) We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. 1) Fin
tekilochka [14]

Answer:

1

p(b) =  \frac{1}{6}

2

p(k) =  \frac{1}{3}

3

P(a) =  \frac{1}{3}

Step-by-step explanation:

Generally when two fair 6-sided dice is rolled the doubles are

(1 1) , ( 2 2) , (3 3) , (4 4) , ( 5 5 ), (6 6)

The total outcome of doubles is N = 6

The total outcome of the rolling the two fair 6-sided dice is

n = 36

Generally the probability that doubles (i.e., having an equal number on the two dice) were rolled is mathematically evaluated as

p(b) =  \frac{N}{n}

p(b) =  \frac{6}{36}

p(b) =  \frac{1}{6}

Generally when two fair 6-sided dice is rolled the outcome whose sum is 4 or less is

(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)

Looking at this outcome we see that there are two doubles present

So

The conditional probability that doubles were rolled is mathematically represented as

p(k) =  \frac{2}{6}

p(k) =  \frac{1}{3}

Generally when two fair 6-sided dice is rolled the number of outcomes that would land on different numbers is L = 30

And the number of outcomes that at least one die is a 1 is W = 10

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The conditional probability that at least one die is a 1 is mathematically represented as

P(a) =  \frac{W}{L}

=> P(a) =  \frac{10}{30}

=> P(a) =  \frac{1}{3}

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i feel like u are going to delete this but if this helped please don't delete it the answer is, Simon is correct because even though the input values are opposite in the reflected function, any real number can be an input.

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icang [17]
We know that
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</span>
the answer is
[12.25,12.35)


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